y= (x-2)/(x^2-x+1)
Find y` and y``
for the problem i get to
y`=(-x^2+4x-1)/(x^2-x+1)^2 by u/v method but the answer of this derivative problem is (Given in the book)
-3(x^2-3x+1)/(x^2-x+1)^3
How can i Get to this form ?
Thanks
$\displaystyle y=\frac{x-2}{x^2-x+1}$
I used the quotient rule (I think you did too):
$\displaystyle \frac{dy}{dx}=\frac{v\frac{du}{dx}-u \frac{dv}{dx}}{v^2}$
where u is the numerator and v is the denominator. In this case:
$\displaystyle u=x-2$ and $\displaystyle v=x^2-x+1$.
So I did it like this:
$\displaystyle \frac{dy}{dx}=\frac{(x^2-x+1)(1)-(x-2)(2x-1)}{(x^2-x+1)^2}$
$\displaystyle \frac{dy}{dx}=\frac{x^2-x+1-2x^2+5x-2}{(x^2-x+1)^2}$
$\displaystyle \frac{dy}{dx}=\frac{-x^2+4x-1}{(x^2-x+1)^2}$
So you're right!
I think it's a typo or just this in another form.
If you put x=1 into both the equations you get two different answers so they're not the same.
The only way to get $\displaystyle (x^2-x+1)^3$ as the denominator is to differentiate again. I thought this would be f''(x).
However, I began to differentiate again (using the quotient rule, it's similar to before) and I got a cubic appearing on top (hence this can't be f''(x)).
Try and differentiate f'(x) again and see what you get. I'm pretty sure i'm doing it correctly.